The extremal of the functional $\int_{0}^{\alpha}{\left((y')^2 - y^2\right)dx}$ that passes through (0,0) and (${\alpha}$,0) has a
- weak minimum if ${\alpha}$ < $\pi$
- strong minimum if ${\alpha}$ < $\pi$
- weak minimum if ${\alpha}$ > $\pi$
- strong minimum if ${\alpha}$ > $\pi$
Comparing the given functional to the standard form
$J = \int_{a}^{b} F(x, y, y^{'})$,
we have $F(x, y, y^{'}) = {\left((y')^2 - y^2\right)}$
and the Euler equation implies that the extremals must satisfy the differential equation $y{''} + y = 0$
Thus, the extremals are given by $y(x)=A\cos(x+\delta_1)+B\sin(x+\delta_2)$.
The boundary condition y(0) = 0 implies A = 0 and $y({\alpha}) = 0$ implies ${\alpha} + {\delta_2} = n{\pi}$ , where n is an integer.
Now , $F_{y'y'}=2>0$ so the function achieves strong minimum.
I am confused how the bound on ${\alpha}$ has to be used.
The correct options are 1 & 2. This question appeared in CSIR Net Exam June 2015.
Thanks in advance!